HSC 2012 MX2 Marathon (archive) (1 Viewer)

Trebla · Mar 13, 2012

Re: 2012 HSC MX2 Marathon

mirakon said:
is the riemann hypothesis thing actually syllabus?

lol, this is a Riemann sum for an integral. The Riemann hypothesis is a completely different thing altogether.

mirakon · Mar 13, 2012

Re: 2012 HSC MX2 Marathon

largarithmic said:
Uh thats not the riemann hypothesis...

Also you're usman, right?

Yes lol, I guess my noobness in maths gave it away haha

deswa1 · Mar 13, 2012

Re: 2012 HSC MX2 Marathon

lolcakes52 said:
Since we are on important mathematical problems, prove P is not NP.

P=NP if P=0 or N=1. So as long as neither of those cases apply, P does not equal NP. Do I get $1,000,000 now?

zeebobDD · Mar 13, 2012

Re: 2012 HSC MX2 Marathon

lolcakes52 said:
Since we are on important mathematical problems, prove P is not NP.

lol like finding both values of the Heisenberg's Uncertainty Principle!::O

Carrotsticks · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

Here's a nice simple one.

Find the locus of:

$|z-i|\geq |z-1|$

Note: Try to use a method that minimises algebra.

nightweaver066 · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Here's a nice simple one.

Find the locus of:

$|z-i|\geq |z-1|$

Note: Try to use a method that minimises algebra.

The region of
$y \leq x$

Carrotsticks · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
The region of
$y \leq x$

How did you come by this answer?

zeebobDD · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

Assume two people are playing a game with a Dice, X and Y, to determine the winner, the player who rolls a 6 first will win the game

If X starts rolling first, Find the probablility that X will win the game

nightweaver066 · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
How did you come by this answer?

|z - i| starts from 1 on the imaginary axis.

|z - 1| starts from 1 on the real axis.

Perpendicular bisector of the chord joining those two points is y = x

For |z - i| >= |z - 1|, has to be the region below and including y = x, therefore $y \leq x$

nightweaver066 · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

zeebobDD said:
Assume two people are playing a game with a Dice, X and Y, to determine the winner, the player who rolls a 6 first will win the game

If X starts rolling first, Find the probablility that X will win the game

P(winning on first go) = 1/6

P(winning on second go) = 5/6 x 5/6 x 1/6

P(winning on third go) = 5/6 x 5/6 x 5/6 x 5/6 x 1/6

By continuing the trend and adding all the possible cases..

P(X winning) = 1/6 + 1/6 x (5/6)^2 + 1/6 x (5/6)^4 + ...

Common ratio of (5/6)^2 therefore it is a geometric progression.

= $\frac{\frac{1}{6}}{1 - (\frac{5}{6})^2}$

$= \frac{6}{11}$

Carrotsticks · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
|z - i| starts from 1 on the imaginary axis.

|z - 1| starts from 1 on the real axis.

Perpendicular bisector of the chord joining those two points is y = x

For |z - i| >= |z - 1|, has to be the region below and including y = x, therefore $y \leq x$

Nice.

The reason why I asked is because as I was waiting for a MATH1001 (Differential Calculus) tutorial, I observed the tutor letting z=x+iy etc and had about half a board of working out before arriving at the same answer, which could have been acquired within a few seconds *after thinking for a bit*.

nightweaver066 · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Nice.

The reason why I asked is because as I was waiting for a MATH1001 (Differential Calculus) tutorial, I observed the tutor letting z=x+iy etc and had about half a board of working out before arriving at the same answer, which could have been acquired within a few seconds *after thinking for a bit*.

lol.. I hope whatever locus you were looking at in your tutorial was more complex than the one you gave.

Carrotsticks · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

Oh sorry, I wasn't clear. I was waiting outside for the first year tutorial to finish.

This subject MATH1001 is mostly done by people who did not do Extension 2 Maths, so the idea of finding locus with complex numbers etc is alien to the vast majority.

Anyway new question, a little bit different from what you guys/girls are used to seeing =)

$\\ $Prove that the following expression is only an integer if n is odd:$ \\\\ \prod_{k=2}^{n} \left (1+ \frac{1}{k} \right )$

Hint: Use the equivalent of a telescoping sum, but for products.

zeebobDD · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
P(winning on first go) = 1/6

P(winning on second go) = 5/6 x 5/6 x 1/6

P(winning on third go) = 5/6 x 5/6 x 5/6 x 5/6 x 1/6

By continuing the trend and adding all the possible cases..

P(X winning) = 1/6 + 1/6 x (5/6)^2 + 1/6 x (5/6)^4 + ...

Common ratio of (5/6)^2 therefore it is a geometric progression.

= $\frac{\frac{1}{6}}{1 - (\frac{5}{6})^2}$

$= \frac{6}{11}$

uhmm why do you multiply it by 5/6 * 5/6 on the second go.. etc

nightweaver066 · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

zeebobDD said:
uhmm why do you multiply it by 5/6 * 5/6 on the second go.. etc

I thought that they rolled the dice one at a time.

For X to win on the second go, he has to miss the first time, Y then has to miss, and then he gets 6 on his next roll.

Apply the same thing for the next go.

Nooblet94 · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

zeebobDD said:
uhmm why do you multiply it by 5/6 * 5/6 on the second go.. etc

Because for X to win the second time, he must lose on his roll (probability 5/6), and Y must also lose (probability 5/6).

EDIT: beat me to it

Nooblet94 · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Oh sorry, I wasn't clear. I was waiting outside for the first year tutorial to finish.

This subject MATH1001 is mostly done by people who did not do Extension 2 Maths, so the idea of finding locus with complex numbers etc is alien to the vast majority.

Anyway new question, a little bit different from what you guys/girls are used to seeing =)

$\\ $Prove that the following expression is only an integer if n is odd:$ \\\\ \prod_{k=2}^{n} \left (1+ \frac{1}{k} \right )$

Hint: Use the equivalent of a telescoping sum, but for products.

<a href="http://www.codecogs.com/eqnedit.php?latex=\prod_{k=2}^n\left(1@plus;\frac{1}{k}\right) = \prod_{k=2}^n\left(\frac{k@plus;1}{k}\right) = \frac{3}{2}\cdot\frac{4}{3}\cdot \frac{5}{4}\cdot ...\cdot \frac{n}{n-1}\cdot\frac{n@plus;1}{n} =\frac{n@plus;1}{2}\\ ~\\ $For the product to be an integer $n@plus;1$ must be divisible by 2.\\ $\therefore n$ must be odd" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\prod_{k=2}^n\left(1+\frac{1}{k}\right) = \prod_{k=2}^n\left(\frac{k+1}{k}\right) = \frac{3}{2}\cdot\frac{4}{3}\cdot \frac{5}{4}\cdot ...\cdot \frac{n}{n-1}\cdot\frac{n+1}{n} =\frac{n+1}{2}\\ ~\\ $For the product to be an integer $n+1$ must be divisible by 2.\\ $\therefore n$ must be odd" title="\prod_{k=2}^n\left(1+\frac{1}{k}\right) = \prod_{k=2}^n\left(\frac{k+1}{k}\right) = \frac{3}{2}\cdot\frac{4}{3}\cdot \frac{5}{4}\cdot ...\cdot \frac{n}{n-1}\cdot\frac{n+1}{n} =\frac{n+1}{2}\\ ~\\ $For the product to be an integer $n+1$ must be divisible by 2.\\ $\therefore n$ must be odd" /></a>

New question:

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ P(x_1,y_1)$ is a point on the ellipse $\frac{x^2}{a^2}@plus;\frac{y^2}{b^2}=1.$ \\~\\A line drawn from the centre $O(0,0)$ parallel to the tangent at $P$ meets the ellipse at $Q$. \\~\\Prove that the area of $\triangle OPQ$ is independent of the position of $P." target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ P(x_1,y_1)$ is a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ \\~\\A line drawn from the centre $O(0,0)$ parallel to the tangent at $P$ meets the ellipse at $Q$. \\~\\Prove that the area of $\triangle OPQ$ is independent of the position of $P." title="\\ P(x_1,y_1)$ is a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ \\~\\A line drawn from the centre $O(0,0)$ parallel to the tangent at $P$ meets the ellipse at $Q$. \\~\\Prove that the area of $\triangle OPQ$ is independent of the position of $P." /></a>

Carrotsticks · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

Haha that question has been asked to death.

But nonetheless it is indeed a good one to practise!

jet · Mar 14, 2012

Re: 2012 HSC MX2 Marathon

Here's a nice one:

$\text{Let } f(x) = x^{\alpha} - \alpha x + \alpha - 1, \text{ where } 0 < \alpha < 1 \\ \\ \text{a) Prove that } f(x) \leq 0 \text{ for } x \geq 0 \\ \\ \text{b Hence, show that } \\ a^{\frac{1}{p}}b^{\frac{1}{q}} \leq \frac{a}{p} + \frac{b}{q} \\ \text{ where } a, b \text{ are non-negative real numbers }, p > 1 \text{ and } \frac{1}{p} + \frac{1}{q} = 1$

kingkong123 · Mar 15, 2012

Re: 2012 HSC MX2 Marathon

New question:

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ P(x_1,y_1)$ is a point on the ellipse $\frac{x^2}{a^2}@plus;\frac{y^2}{b^2}=1.$ \\~\\A line drawn from the centre $O(0,0)$ parallel to the tangent at $P$ meets the ellipse at $Q$. \\~\\Prove that the area of $\triangle OPQ$ is independent of the position of $P." target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ P(x_1,y_1)$ is a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ \\~\\A line drawn from the centre $O(0,0)$ parallel to the tangent at $P$ meets the ellipse at $Q$. \\~\\Prove that the area of $\triangle OPQ$ is independent of the position of $P." title="\\ P(x_1,y_1)$ is a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ \\~\\A line drawn from the centre $O(0,0)$ parallel to the tangent at $P$ meets the ellipse at $Q$. \\~\\Prove that the area of $\triangle OPQ$ is independent of the position of $P." /></a>

$P(acos\theta,bsin\theta)\\\therefore m_{tangent}=\frac{bcos\theta}{-asin\theta}\\\therefore $Equation of OQ $\Rightarrow y=-\frac{bcos\theta x}{asin\theta}\\$Sub into $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and you get $ y=\pm asin\theta $ $ \therefore x=\mp bcos\theta\\\\$ Take one of the values $ \therfore Q(asin\theta,-bcos\theta)\\$Now use perpendicular distance formula from $ P$ to the line $ OQ. \\\\\perp D=\frac{ab}{\sqrt{b^2cos^2\theta+a^2sin^2\theta}}\\\\$Now use distance forumla to find $ OQ \\\\|OQ|=\sqrt{b^2cos^2\theta+a^2sin^2\theta}\\\\ \therefore $ Area $\Delta OPQ=\frac{1}{2}\times\frac{ab\sqrt{b^2cos^2\theta+a^2sin^2\theta}}{\sqrt{b^2cos^2\theta+a^2sin^2\theta}}\\=\frac{1}{2}ab$ which is independent of $\theta$ and hence independent of the position of P$$

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